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Soluzione del esercizio 157 di Matematica.verde, Vol. 1 (seconda edizione)
Esercizi svolti per il libro Matematica.verde, Vol. 1 (seconda edizione)
Capitolo
Scegli il capitolo
Capitolo 6 - I monomi
Capitolo 6 - Allenamento
Capitolo 6 - Prova B
Capitolo 7 - I polinomi
Esercizio
Scegli l'esercizio
96
97
98
99
37
146
147
148
149
150
156
157
202
201
233
234
235
236
237
174
204
241
242
243
244
245
246
247
209
217
259
260
264
265
266
275
277
278
283
293
313
369
371
372
378
401
Soluzione
(
2
a
+
3
b
)
(
−
2
b
⋅
−
1
3
a
)
−
(
3
4
a
b
⋅
2
a
+
3
4
a
b
⋅
b
)
−
(
a
b
⋅
1
2
a
+
a
b
⋅
3
4
b
)
(2a+3b)(-2b \cdot -\dfrac{1}{3}a)-(\dfrac{3}{\cancel4}ab \cdot \cancel2a + \dfrac{3}{4}ab \cdot b) - (ab \cdot \dfrac{1}{2}a + ab \cdot \dfrac{3}{4}b)
(
2
a
+
3
b
)
(
−
2
b
⋅
−
3
1
a
)
−
(
4
3
ab
⋅
2
a
+
4
3
ab
⋅
b
)
−
(
ab
⋅
2
1
a
+
ab
⋅
4
3
b
)
(
2
a
+
3
b
)
(
2
3
a
b
)
−
(
3
2
a
2
b
+
3
4
a
b
2
)
−
(
1
2
a
2
b
+
3
4
a
b
2
)
(2a+3b)(\dfrac{2}{3}ab) - (\dfrac{3}{2}a^2b+\dfrac{3}{4}ab^2) - (\dfrac{1}{2}a^2b + \dfrac{3}{4}ab^2)
(
2
a
+
3
b
)
(
3
2
ab
)
−
(
2
3
a
2
b
+
4
3
a
b
2
)
−
(
2
1
a
2
b
+
4
3
a
b
2
)
(
2
a
⋅
2
3
a
b
+
3
b
⋅
2
3
a
b
)
−
3
2
a
2
b
−
3
4
a
b
2
−
1
2
a
2
b
−
3
4
a
b
2
(2a \cdot \dfrac{2}{3}ab + \cancel3b \cdot \dfrac{2}{\cancel3}ab) - \dfrac{3}{2}a^2b -\dfrac{3}{4}ab^2 -\dfrac{1}{2}a^2b - \dfrac{3}{4}ab^2
(
2
a
⋅
3
2
ab
+
3
b
⋅
3
2
ab
)
−
2
3
a
2
b
−
4
3
a
b
2
−
2
1
a
2
b
−
4
3
a
b
2
4
3
a
2
b
+
2
a
b
2
−
3
2
a
2
b
−
3
4
a
b
2
−
1
2
a
2
b
−
3
4
a
b
2
\dfrac{4}{3}a^2b + 2ab^2 - \dfrac{3}{2}a^2b - \dfrac{3}{4}ab^2 - \dfrac{1}{2}a^2b - \dfrac{3}{4}ab^2
3
4
a
2
b
+
2
a
b
2
−
2
3
a
2
b
−
4
3
a
b
2
−
2
1
a
2
b
−
4
3
a
b
2
(
4
3
a
2
b
−
3
2
a
2
b
−
1
2
a
2
b
)
+
(
2
a
b
2
−
3
4
a
b
2
−
3
4
a
b
2
)
(\dfrac{4}{3}a^2b - \dfrac{3}{2}a^2b - \dfrac{1}{2}a^2b) + (2ab^2 - \dfrac{3}{4}ab^2 - \dfrac{3}{4}ab^2)
(
3
4
a
2
b
−
2
3
a
2
b
−
2
1
a
2
b
)
+
(
2
a
b
2
−
4
3
a
b
2
−
4
3
a
b
2
)
(
8
a
2
b
−
9
a
2
b
−
3
a
2
b
6
)
+
(
8
a
b
2
−
3
a
b
2
−
3
a
b
2
4
)
(\dfrac{8a^2b-9a^2b-3a^2b}{6})+(\dfrac{8ab^2 - 3ab^2 - 3ab^2}{4})
(
6
8
a
2
b
−
9
a
2
b
−
3
a
2
b
)
+
(
4
8
a
b
2
−
3
a
b
2
−
3
a
b
2
)
−
4
a
2
b
6
+
2
a
b
2
4
\dfrac{-\cancel4a^2b}{\cancel6} + \dfrac{\cancel2ab^2}{\cancel4}
6
−
4
a
2
b
+
4
2
a
b
2
−
2
3
a
2
b
+
1
2
a
b
2
-\dfrac{2}{3}a^2b + \dfrac{1}{2}ab^2
−
3
2
a
2
b
+
2
1
a
b
2